---
title: "Log amplifier transfer function, matched transistors"
subject: "Advanced Electronics"
chapter: 5
key: "semester_4/advanced_electronics:log-amp-matched-transfer"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/log-amp-matched-transfer
site: BCT Engineering
---

# Log amplifier transfer function, matched transistors

Advanced Electronics · Chapter 5: Op-Amp Bipolar Transistor Logarithmic Amplifier

- Asked in: 2071 Shrawan, 5 marks; 2070 Chaitra, 5 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/log-amp-matched-transfer
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-5

## Question

Draw the circuit for the logarithmic amplifier using matched transistor and derive its transfer function.

## Short answer

Two log stages with matched transistors produce Vo1 = -VT ln(Vin/(R1*Is)) and Vo2 = -VT ln(Vref/(R2*Is)). The difference amplifier subtracts them: Vo = (R4/R3) VT ln((Vin R2)/(Vref R1)). With R1=R2, Vo = (R4/R3) VT ln(Vin/Vref); Is cancels.

## Answer

A log amplifier gives an output proportional to the logarithm of its input. In the matched-transistor version, a second identical transistor carrying a reference current cancels the saturation current Is, which depends strongly on temperature.

#### Circuit
```text
 Vin-[R1]-+-(-)\   Q1: C at (-), E at Vo1
          |  A1 >--Vo1--[R3]--+--[R4]--+
 GND-----(+)/                 |        |
                             (-)\      |
                                 A3>---+--> Vo
                             (+)/
 Vref-[R2]-+-(-)\   Q2        |
           |  A2 >--Vo2--[R3]-+--[R4]--GND
 GND------(+)/
 (bases of Q1, Q2 grounded; Q1, Q2 matched)
```

A1 and A2 are log stages. A3 is a difference amplifier of gain R4/R3.

#### Derivation
The transistor law gives VBE = VT ln(IC/Is).

**Stage A1:**
```text
IC1 = Vin/R1
Vo1 = −VBE1 = −VT ln( Vin / (R1·Is) )
```

**Stage A2:**
```text
IC2 = Vref/R2
Vo2 = −VBE2 = −VT ln( Vref / (R2·Is) )
```

**Difference amplifier A3:**
```text
Vo = (R4/R3)(Vo2 − Vo1)
   = (R4/R3)·VT [ ln(Vin/(R1·Is)) − ln(Vref/(R2·Is)) ]
   = (R4/R3) · VT · ln( (Vin·R2) / (Vref·R1) )
```

#### Transfer function
With R1 = R2:

```text
Vo = (R4/R3) · VT · ln( Vin / Vref )
   = 2.303 (R4/R3) VT · log10( Vin / Vref )
```

- **Is has cancelled**, because Q1 and Q2 are matched and at the same temperature.
- At 300 K, VT ≈ 25.9 mV. With R4/R3 = 1, the output changes by about 60 mV per decade of Vin.
- When Vin = Vref, the output is 0.
- Vin and Vref must be positive (for npn transistors).
- VT = kT/q still varies with temperature. It is compensated by making R3 a temperature-sensitive resistor.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why does the matched-transistor log amp have a second transistor carrying a reference current?
   A. To cancel the temperature-dependent saturation current Is
   B. To increase the output voltage by 60 mV per decade
   C. To allow negative input voltages
   D. To remove VT from the output
   Answer: A. Both transistors have the same Is, so subtracting their log outputs removes Is. (Key point: Purpose of Q2 branch)

2. What is the output of A3 with R1 = R2?
   A. Vo = (R4/R3) VT ln(Vin Vref)
   B. Vo = (R3/R4) VT ln(Vin/(R1 Is))
   C. Vo = -VT ln(Vin/(R1 Is))
   D. Vo = (R4/R3) VT ln(Vin/Vref)
   Answer: D. A3 gives (R4/R3)(Vo2 - Vo1), and the Is terms cancel, leaving the log of the ratio Vin/Vref. (Key point: Difference amplifier result)

3. What is the output when Vin = Vref (R1 = R2)?
   A. Vref
   B. Infinity
   C. 0 V
   D. VT
   Answer: C. ln(1) = 0, so Vo = 0. (Key point: Output at Vin = Vref)

4. At 300 K with R4/R3 = 1, how much does the output change per decade of Vin?
   A. About 600 mV
   B. About 1 V
   C. About 60 mV
   D. About 26 mV
   Answer: C. 2.303 VT = 2.303 x 25.9 mV, which is about 60 mV per decade. (Key point: Sensitivity)

## Related in Advanced Electronics

- [Antilog amplifier with matched transistors](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aantilog-matched-transistor-transfer) — Advanced Electronics, chapter 5, asked 2×
- [Matched-transistor log amp and temperature effects](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Alog-amp-matched-temperature) — Advanced Electronics, chapter 5
- [Basic logarithmic amplifier circuit](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Abasic-log-amplifier) — Advanced Electronics, chapter 5
- [Matched-transistor antilog amp and temperature](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aantilog-matched-temperature) — Advanced Electronics, chapter 5
- [Output of log-ratio amplifier with RT compensation](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Alog-ratio-temperature-compensated) — Advanced Electronics, chapter 5
- [Log-antilog circuit: Vo in terms of Vi and VR](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Alog-antilog-circuit-vo-vi) — Advanced Electronics, chapter 5, asked 3×

## Related in other subjects

- [Transfer function of op-amp phase-shift circuit](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-opamp-allpass-r1-c-r2) — Control Systems, chapter 2, asked 2×
- [Transfer function of op-amp integrator-type circuit](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-opamp-r1-r2-c) — Control Systems, chapter 2
- [Tow-Thomas band pass biquad transfer function](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atow-thomas-bandpass-derivation) — Filter Design, chapter 6
- [Common base amplifier circuit and gains](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Acb-amplifier-analysis) — Electronics Devices and Circuit, chapter 2